√896z^15/<span>√225z^6
= </span>√896z^9/<span>√225
= </span>√64√14(z^4)<span>√z/15
= 8</span>√14(z^4)<span>√z/15
= (8z^4/15) * </span><span>√14z
x = 8</span>
We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.
First of all, we will find z-score corresponding to 38 and 56.


Now we will find z-score corresponding to 56.

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is
.
We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.
We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
Answer:
64
Step-by-step explanation:
x + 5(x + 2)
Distribute
x + 5x+10
Combine like terms
6x+10
Let x = 9
6(9) +10
54+10
64
If you don30,000 + 3,500 you get 33,500 so the answer is 33,500☺️